According to the law of conservation of mechanical energy, the kinetic energy of the moving mass is completely converted into the elastic potential energy of the spring at the point of maximum compression . Therefore, $K_f + V_f = K_i + V_i$ $0 + \frac{1}{2}kx_m^2 = \frac{1}{2}mv^2 + 0$ $\frac{1}{2}kx_m^2 = \frac{1}{2}mv^2$ Where $m = 0.5 \text{ kg}$, $v = 1.5 \text{ m/s}$, and $k = 50 \text{ N/m}$. Solving for $x_m$ (maximum compression): $x_m^2 = \frac{m}{k}v^2$ $x_m^2 = \frac{0.5}{50} \times (1.5)^2$ $x_m^2 = \frac{1}{100} \times 2.25 = 0.0225$ $x_m = \sqrt{0.0225} = 0.15 \text{ m}$ Thus, the maximum compression of the spring is $0.15 \text{ m}$.